Problem: Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:

  * Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). 
  * Digits are written one by one in order corresponding to number and separated by single '0' character. 



Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.

Input

The first line contains one integer number n (1 ≤ n ≤ 89) — length of the string s.

The second line contains string s — sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.

Output

Print the decoded number.

Examples

Input

3
111


Output

3


Input

9
110011101


Output

2031

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void readline(char *s) {
  int len = 0;
  char c;
  for (c = ((char)fgetc(stdin)); c != '\n';
       s[len++] = c, c = ((char)fgetc(stdin)))
    ;
  s[len] = 0;
}
void readstr(char *s) {
  int len = 0;
  char c;
  for (c = ((char)fgetc(stdin)); c != ' ' && c != '\n';
       s[len++] = c, c = ((char)fgetc(stdin)))
    ;
  s[len] = 0;
}
void printstr(const char *s) {
  int len = strlen(s);
  for (int(i) = (0); (i) <= (len - 1); i++) fputc((s[i]), stdout);
}
void to_lower(char *s, int L, int R) {
  for (int(i) = (L); (i) <= (R); i++)
    if (((s[i]) >= 'A' && (s[i]) <= 'Z')) s[i] = s[i] - 'A' + 'a';
}
void to_upper(char *s, int L, int R) {
  for (int(i) = (L); (i) <= (R); i++)
    if (((s[i]) >= 'a' && (s[i]) <= 'z')) s[i] = s[i] - 'a' + 'A';
}
long long readll() {
  char c;
  long long ans = 0, x = 1;
  for (c = ((char)fgetc(stdin)); !((c) >= '0' && (c) <= '9') && c != '-';
       c = ((char)fgetc(stdin)))
    ;
  if (c == '-') x = -1, c = ((char)fgetc(stdin));
  for (; ((c) >= '0' && (c) <= '9');
       ans = ans * 10 + c - '0', c = ((char)fgetc(stdin)))
    ;
  return ans * x;
}
void printll(long long x) {
  int a[25], cnt = 0;
  if (x == 0) {
    fputc(('0'), stdout);
    return;
  }
  if (x < 0) fputc(('-'), stdout), x = -x;
  for (; x; a[cnt++] = x - x / 10 * 10, x /= 10)
    ;
  for (int(i) = (cnt - 1); (i) >= (0); i--) fputc((a[i] + '0'), stdout);
}
int n, x[105];
int main() {
  n = ((int)readll());
  for (int(i) = (1); (i) <= (n); i++) x[i] = ((char)fgetc(stdin)) - '0';
  for (int(i) = (1); (i) <= (n + 1); i++) {
    if (x[i] == 0) {
      int j = i - 1;
      while (j && x[j] == 1) j--;
      if (x[j] == 0) printll((long long)(i - j - 1));
    }
  }
  return 0;
}